Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{-9x^2 + 36x}{x^2 - 7x + 12} \times \dfrac{5x - 15}{9x - 54} $
Solution: First factor the quadratic. $n = \dfrac{-9x^2 + 36x}{(x - 3)(x - 4)} \times \dfrac{5x - 15}{9x - 54} $ Then factor out any other terms. $n = \dfrac{-9x(x - 4)}{(x - 3)(x - 4)} \times \dfrac{5(x - 3)}{9(x - 6)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -9x(x - 4) \times 5(x - 3) } { (x - 3)(x - 4) \times 9(x - 6) } $ $n = \dfrac{ -45x(x - 4)(x - 3)}{ 9(x - 3)(x - 4)(x - 6)} $ Notice that $(x - 4)$ and $(x - 3)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -45x(x - 4)\cancel{(x - 3)}}{ 9\cancel{(x - 3)}(x - 4)(x - 6)} $ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ $n = \dfrac{ -45x\cancel{(x - 4)}\cancel{(x - 3)}}{ 9\cancel{(x - 3)}\cancel{(x - 4)}(x - 6)} $ We are dividing by $x - 4$ , so $x - 4 \neq 0$ Therefore, $x \neq 4$ $n = \dfrac{-45x}{9(x - 6)} $ $n = \dfrac{-5x}{x - 6} ; \space x \neq 3 ; \space x \neq 4 $